a) \(n_{K2SO4}=\dfrac{17,4}{174}=0,1\left(mol\right)\)
PTHH : \(K_2SO_4+BaCl_2-->BaSO_4\downarrow+2KCl\)
Theo PTHH :nBaSO4 = nK2SO4 = 0,1 (mol)
=> mBaSO4 = 0,1. 233 = 23,3 (g)
b) Theo PTHH :
nKCl = 2nK2SO4 = 0,2 (mol)
nBaCl2 = nK2SO4 = 0,1 (mol)
=> mBaCl2 = 0,1.208 = 20,8 (g)
=> m(ddBaCl2) = 20,8 : 10.100 = 208 (g)
Áp dụng định luật bảo toàn khối lượng :
mK2SO4 + m(ddBaCl2) = mBaSO4 + m(ddKCl)
=> 17,4 + 208 = 23,3 + m(ddKCl)
=> m(ddKCl) = 202,1 (g)
=> \(C\%KCl=\dfrac{0,2.74,5}{202,1}\cdot100\%\approx7,37\%\)
\(\begin{array}{l} a,\\ n_{K_2SO_4}=\dfrac{17,4}{174}=0,1\ (mol)\\ PTHH:K_2SO_4+BaCl_2\to BaSO_4\downarrow+2KCl\\ Theo\ pt:\ n_{BaSO_4}=n_{K_2SO_4}=0,1\ (mol)\\ \Rightarrow m_{BaSO_4}=0,1\times 233=23,3\ (g)\\ b,\\ Theo\ pt:\ n_{BaCl_2}=n_{K_2SO_4}=0,1\ (mol)\\ \Rightarrow m_{\text{dd BaCl_2}}=\dfrac{0,1\times 208}{10\%}=208\ (g)\\ m_{\text{dd spư}}=m_{K_2SO_4}+m_{\text{dd BaCl_2}}-m_{BaSO_4}\\ \Rightarrow m_{\text{dd spư}}=17,4+208-23,3=202,1\ (g)\\ Theo\ pt:\ n_{KCl}=2n_{K_2SO_4}=0,2\ (mol)\\ \Rightarrow C\%_{\text{dd spư}}=C\%_{KCl}=\dfrac{0,2\times 74,5}{202,1}\times 100\%=7,37\%\end{array}\)
